具有至少一个公共点的重叠矩形的最大数量
给定四个数组x1[] , x2[] , y1[] , y2[]大小为N其中 ( x1[i] , y1[i] ) 表示左下角并且 ( x2[i] , y2[i] ) 矩形的右上角，任务是找到至少有一个公共点的重叠矩形的最大数量。
例子：
Input: N = 2 x1[] = {0, 50} y1[] = {0, 50} x2[] = {100, 60} y2[] = {100, 60}
Output: 2
Explanation: There are two rectangles {{0, 0}, {100, 100}}, {{50, 50}, {60, 60}} and both of them are intersecting as shown in the given figure below:
Input: N = 2 x1[] = {0, 100} y1[] = {0, 100} x2[] = {100, 200} y2[] = {100, 200}
Output: 1
方法：给定的问题可以通过使用贪心方法来解决，该方法基于以下思想：坐标平面上的每个矩形都有自己的区域，当在同一平面上添加多个矩形时，它们会在彼此之间产生交集。因此，要选择在公共区域上重叠的最大矩形数量，贪婪地选择1×1 单位的区域，因为所有重叠区域都至少有这么多块。请按照以下步骤解决给定的问题：
- 因为有N个矩形，每个矩形有2 个 X 坐标和2 个 Y 坐标。总共将有2*N个X -coordinates和Y-coordinates 。
- 因此，创建一个X和Y坐标向量，并将所有X和Y从各自向量中的矩形推入。
- 初始化一个变量，例如maxRectangles为0 ，它存储最大重叠矩形。
- 遍历向量X[]并且对于每个坐标X[i] ，遍历向量Y[]并找到与X[i]重叠的矩形的数量，在这一步之后，将maxRectangles的值更新为maxRectangles的最大值并计数为当前迭代获得。
- 完成上述步骤后，打印maxRectangles的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the maximum number
// of overlapping rectangles
void maxOverlappingRectangles(
int x1[], int y1[], int x2[],
int y2[], int N)
{
// Stores the maximum count of
// overlapping rectangles
int max_rectangles = 0;
// Stores the X and Y coordinates
vector X, Y;
for (int i = 0; i < N; i++) {
X.push_back(x1[i]);
X.push_back(x2[i] - 1);
Y.push_back(y1[i]);
Y.push_back(y2[i] - 1);
}
// Iterate over all pairs of Xs and Ys
for (int i = 0; i < X.size(); i++) {
for (int j = 0; j < Y.size(); j++) {
// Store the count for the
// current X and Y
int cnt = 0;
for (int k = 0; k < N; k++) {
if (X[i] >= x1[k]
&& X[i] + 1 <= x2[k]
&& Y[j] >= y1[k]
&& Y[j] + 1 <= y2[k]) {
cnt++;
}
}
// Update the maximum count of
// rectangles
max_rectangles = max(
max_rectangles, cnt);
}
}
// Returns the total count
cout << max_rectangles;
}
// Driver Code
int main()
{
int x1[] = { 0, 50 };
int y1[] = { 0, 50 };
int x2[] = { 100, 60 };
int y2[] = { 100, 60 };
int N = sizeof(x1) / sizeof(x1[0]);
maxOverlappingRectangles(
x1, y1, x2, y2, N);
return 0;
} Java
// java program for the above approach
import java.util.*;
class GFG
{
// Function to find the maximum number
// of overlapping rectangles
static void maxOverlappingRectangles(int[] x1, int[] y1,
int[] x2, int[] y2,
int N)
{
// Stores the maximum count of
// overlapping rectangles
int max_rectangles = 0;
// Stores the X and Y coordinates
Vector X = new Vector<>();
Vector Y = new Vector<>();
for (int i = 0; i < N; i++) {
X.add(x1[i]);
X.add(x2[i] - 1);
Y.add(y1[i]);
Y.add(y2[i] - 1);
}
// Iterate over all pairs of Xs and Ys
for (int i = 0; i < X.size(); i++) {
for (int j = 0; j < Y.size(); j++) {
// Store the count for the
// current X and Y
int cnt = 0;
for (int k = 0; k < N; k++) {
if (X.get(i)>= x1[k] && X.get(i) + 1 <= x2[k]
&& Y.get(j) >= y1[k]
&& Y.get(j) + 1 <= y2[k]) {
cnt++;
}
}
// Update the maximum count of
// rectangles
max_rectangles
= Math.max(max_rectangles, cnt);
}
}
// Returns the total count
System.out.println(max_rectangles);
}
// Driver Code
public static void main(String[] args)
{
int[] x1 = { 0, 50 };
int[] y1 = { 0, 50 };
int[] x2 = { 100, 60 };
int[] y2 = { 100, 60 };
int N = x1.length;
maxOverlappingRectangles(x1, y1, x2, y2, N);
}
}
// This code is contributed by amreshkumar3. Python3
# Python3 program for the above approach
# Function to find the maximum number
# of overlapping rectangles
def maxOverlappingRectangles(x1, y1, x2, y2, N):
# Stores the maximum count of
# overlapping rectangles
max_rectangles = 0
# Stores the X and Y coordinates
X = []
Y = []
for i in range(0, N):
X.append(x1[i])
X.append(x2[i] - 1)
Y.append(y1[i])
Y.append(y2[i] - 1)
# Iterate over all pairs of Xs and Ys
for i in range(0, len(X)):
for j in range(0, len(Y)):
# Store the count for the
# current X and Y
cnt = 0
for k in range(0, N):
if (X[i] >= x1[k] and X[i] + 1 <= x2[k] and Y[j] >= y1[k] and Y[j] + 1 <= y2[k]):
cnt += 1
# Update the maximum count of
# rectangles
max_rectangles = max(max_rectangles, cnt)
# Returns the total count
print(max_rectangles)
# Driver Code
if __name__ == "__main__":
x1 = [0, 50]
y1 = [0, 50]
x2 = [100, 60]
y2 = [100, 60]
N = len(x1)
maxOverlappingRectangles(x1, y1, x2, y2, N)
# This code is contributed by rakeshsahniC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the maximum number
// of overlapping rectangles
static void maxOverlappingRectangles(int[] x1, int[] y1,
int[] x2, int[] y2,
int N)
{
// Stores the maximum count of
// overlapping rectangles
int max_rectangles = 0;
// Stores the X and Y coordinates
List X = new List();
List Y = new List();
for (int i = 0; i < N; i++) {
X.Add(x1[i]);
X.Add(x2[i] - 1);
Y.Add(y1[i]);
Y.Add(y2[i] - 1);
}
// Iterate over all pairs of Xs and Ys
for (int i = 0; i < X.Count; i++) {
for (int j = 0; j < Y.Count; j++) {
// Store the count for the
// current X and Y
int cnt = 0;
for (int k = 0; k < N; k++) {
if (X[i] >= x1[k] && X[i] + 1 <= x2[k]
&& Y[j] >= y1[k]
&& Y[j] + 1 <= y2[k]) {
cnt++;
}
}
// Update the maximum count of
// rectangles
max_rectangles
= Math.Max(max_rectangles, cnt);
}
}
// Returns the total count
Console.WriteLine(max_rectangles);
}
// Driver Code
public static void Main()
{
int[] x1 = { 0, 50 };
int[] y1 = { 0, 50 };
int[] x2 = { 100, 60 };
int[] y2 = { 100, 60 };
int N = x1.Length;
maxOverlappingRectangles(x1, y1, x2, y2, N);
}
}
// This code is contributed by ukasp. Javascript
输出：
2时间复杂度： O(N 3 )
辅助空间： O(N)